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Electrons absorbing light?

  1. Jan 1, 2012 #1
    Sorry I posted 2 threads within the hour, but I'm just really interested. In my chem class they (teachers, txtbooks, etc.) always mention how electrons "absorb" photons and "jump" to higher energy levels, but when I was reading about QED, renormalization, compton scattering, etc. they talk about how the photons are deflected when they hit an electron and they treat the situation more classically. What is a reason(s) for this?
     
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  3. Jan 1, 2012 #2

    morrobay

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    Compton scattering is from x-ray radiation and is described with photons that have
    "billiard- ball " like collisions with the electrons in the medium. And these recoiled photons
    are scattered.
    Where as the natural resonate frequency for electrons in atoms and molecules corresponds
    to the violet and ultraviolet region. Absorption, reflection, scattering, refraction are described in terms of a wave.
    Where the time varying electric vector of the incident wave causes the electrons in
    the medium to oscillate.
     
    Last edited: Jan 1, 2012
  4. Jan 1, 2012 #3
    Ok. But why is it that a bound electron absorbs a photon whereas a free electron does not
     
  5. Jan 2, 2012 #4
    Conservation of energy/momentum. A free electron's four-momentum must always have a magnitude equal to its rest mass. Therefore, it's impossible for electron + photon and excited-electron to have the same total four-momentum. However, in a bound state, the electron can have a range of different energies, so it is possible for those two things to have the same four-momentum.
     
  6. Jan 2, 2012 #5

    morrobay

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    An electron in an atom or molecule is bound there by strong restoring forces.
    It has a definite natural frequency.
     
  7. Jan 2, 2012 #6
    I kind of get what you are saying. So is this just one of those guage invariance situations where nature has to compensate for itself if you know what I mean. Im not sure if im getting the whole 4 momentum thing either
     
  8. Jan 2, 2012 #7
    Oh so a bound electron can acquire some sort of vibration and if that matches the photon's frequency it will be absorbed? I know about how molecules/atoms can vibrate and resonate with certain light frequencies which allow them to absorb the light, would that be the same application?
     
  9. Jan 2, 2012 #8
    Sort of. The electron doesn't really "vibrate" in the classical sense, but when it's in the potential well of a nucleus, there are a different set of energy states that it's allowed to be in, which it wouldn't be allowed to be in if it were free.

    Think of it this way: an electron starts out with energy [itex]E[/itex], and the photon has energy [itex]E_p[/itex]. If the electron absorbs the photon, conservation of energy says that it must now have energy [itex]E + E_p[/itex]. For a free electron, a state with this energy is not allowed, so it's impossible for the electron to absorb the photon. For a bound electron, though, that state is available, so it is possible for it to absorb the photon. The real physics are a bit more complicated than that (they involve some relativity things that connect energy and momentum), but that's the basic idea.
     
  10. Jan 2, 2012 #9

    tom.stoer

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    In an atom it's not the electron that absorbes the energy of the incoming photon but the whole atom; the usual QM descritpion of the hydrogen atom is a bit misleading here b/c it treats the proton classically, but it should be clear that a more realistic picture is a two-particle Schrödinger equation where the proton-electron system as whole can absorb the photon whereas a single, free electron can't due to energy-momentum conservation.

    In order to absorb a photon there must be an inelastic collision which is impossible w/o additonal degrees of freedom that can be excited; in our case these degrees of freedom are just the "vibrations" of the electron-proton-system; the electron alone as an elementary particle does not have such additonal degrees of freedom so it can't absorb the photon.
     
  11. Jan 2, 2012 #10
    Hmm, that's different than my understanding of the situation. I'm still learning, so I'll certainly trust your view over mine, but now I have a question of my own. Even in normal QM, the existence of a bound state makes possible a set of energy levels that wasn't there for a free particle, even when we treat the proton as just a classical potential field. We can also come up with some sort of notion of energy level transitions in QM, however ham-handed, just by talking about a time-dependent perturbation to the Hamiltonian. This would make me assume that we're justified in talking about the electron states in some sensible way without involving the exact details of the proton.

    Say we now move to QFT, and treat the electron as a full QFT system (Feynman diagrams, creation/annihilation ops, etc.) but still treat the proton as a classical potential term in the Lagrangian. It should still be possible to describe the electron as forming bound states in this scenario, right? They should look like the normal QM states, but with relativistic corrections. You're saying that it's not possible for this model to describe the process of absorbing/emitting a photon and carrying the electron to a higher energy state, even though we're now describing the electron with a full field theory? I'm not exactly clear how QFT handles bound states yet, so while I thought it was possible to do this, it sounds like that's actually wrong?
     
  12. Jan 2, 2012 #11

    tom.stoer

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    There's no need to use QFT.

    The basic observation is that inelastic scattering is not compatible with elementary particles w/o additional degrees of freedom. In our case the additional degrees of freedom are provided by the substructure of the proton-electron system, i.e. the energy levels of this system. Again even in the atom it's not the electron that absorbs the photon, it's the whole atom; otherwise energy-monmentum conservation would be violated.
     
  13. Jan 2, 2012 #12
    Ok, let me see if I understand correctly. The state of the electron before absorption can be described by modelling the proton as a simple potential field. The state after absorption can also be modelled that way--these are just the hydrogen atom energy states that everybody calculates in basic QM. If I understand you right, what you're saying is that since an energy transition in QM is modelled as a time-dependent perturbation to the Hamiltonian, we need something that can wiggle the shape of the potential well in order to coax our electron up into a higher state, correct? So we have to think of the proton as vibrating around in some sense, to create a Hamiltonian that changes in time, so that it can impart energy to the electron.

    So in summary, while we can describe the steady states of the atom by simply ignoring the details of the proton, during absorption/emission we have to talk about both pieces wiggling around simultaneously in order for it to make sense?
     
  14. Jan 3, 2012 #13

    tom.stoer

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    No, you didn't understand correctly.

    What I am saying is this: for a particle to absorb a photon you need inzernal degrees of freedom which can be excited. A free electron can't absorb a photon due to the non-existence of these inner degrees of freedom. An electron bound in an atom can b/c the whole atom (proton-electron bound state) provides these inner degrees of freedom.

    In a rather simple calculation you will see that in order to absorb a photon, the atom has to carry away energy and momentum. The energy is stored in the excitation of the electron-positron bound state (plus recoil); for the momentum you need to take the recoil of the whole atom into account.

    Think about a singe photon which excites a hydrogen atom from 1s to 2p; where does the momentum of the photon go?
     
  15. Jan 3, 2012 #14
    When I first read Chopin's post #4 I didn't get it....but with tom.stoer's explanation maybe I do now.....

    So are we concluding that this part of Chopin's post #4 :

    is correct...by understanding that it's the system of particles (the electron-proton bound state of the atom) that provides the required additional degrees of freedom and that the
    conservation involved is from this perspective:

    http://en.wikipedia.org/wiki/Four-momentum

    those ideas now seem to fit together.
     
  16. Jan 3, 2012 #15
    If you want, you can interpret Compton scattering as the electron absorbs a photon and immediately emits a photon with a lower energy. And this interpretation is based on the perturbation calculation using creation&annihilation operators. However keep in mind we can't testify this intermediate process, all we can measure is initial and final states.
     
  17. Jan 3, 2012 #16

    tom.stoer

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    Regarding Compton scattering: if you believe in the intermediate state where the electron has 'absorbed' the photon and became a 'virtual electron' there's still energy-momentum conservation, but m² = E²-p² is violated, the virtual electron is 'off-shell'.

    But as kof9595995 said, this intermediate state or virtual particle is not accessable to any experiment.
     
  18. Jan 3, 2012 #17

    Ken G

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    Another point to bear in mind is that it is not necessary for the electron to be bound to the proton for the system to absorb a photon-- in plasmas, the electron can be free, but still in the field of a proton. This is called "free-free absorption", or "inverse bremsstrahlung." Also, there is often a lot of ambiguity in the meaning of the term "absorb". Usually it means the photon is destroyed, so it isn't used for elastic scattering of the photon, yet even elastic scattering can be framed as absorbing one photon and creating a new one (perhaps connected by the "virtual" phase mentioned above), so you really have to look closely at the context of how the word "absorb" is being used-- there really isn't a uniform convention there.
     
  19. Jan 3, 2012 #18

    tom.stoer

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    what I Mean by "absorb" is that there is an initial state "photon + particle" and a final (physical) state "particle" w/o a photon; this is only possible if the particle belongs (and interacts with) a larger system which can absorb energy and/or momentum
     
  20. Jan 4, 2012 #19
    Hmm that is interesting. Ok I see why the photon is absorbed, but what does "absorbed" mean anyway?

    Edit: sorry I did not see you last post. But I am still confused about how the photon is lost in the system.
     
  21. Jan 4, 2012 #20

    Ken G

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    I'm not sure there is any answer to "how the photon is lost". How does something move, or exert forces on something else? We observe phenomena, and try to come up with predictive and descriptive theories, but at some point all theories come down to "that's just what happens." What do you want to know about how the photon is lost?

    One way of looking at it is, the photon induces reactions that create new sources of amplitudes that interfere with the photon amplitude so completely there there is no longer any net photon amplitude going forward. That's not really right either, because photons are indistinguishable so all we can really say is that the occupation number drops by 1. So that can be viewed as the photon stimulating a process that creates, in effect, a -1 photon. The photon is responsible, then, for its own destruction, in a sense, but it requires the participation of not only an electron, but as you heard above, a system for that electron to interact with.

    If you take this view, you can say that to get absorption, two things need to happen. First, the response of the electron must cancel the amplitude of the incident photon, and either free electrons or electrons in conservative potentials can do that (if in a harmonic potential, they will be way better at doing this at the resonant harmonic frequency, and way worse at other frequencies). But to call it absorption, you also have to not create a new outbound amplitude for that photon-- otherwise we'd call it scattering instead. That's where you need some additional complexity-- and a simple conservative potential, even one that binds the electron to an atom, is not enough. There needs to be some competing process, something that can interfere with the creation of an outbound amplitude, such that the photon can actually be destroyed (and its energy shunted over into energy associated with the competing process). That often requires a second electron in the vicinity, which can take up the lost energy (often a free electron, which we then call "collisional destruction" when the free electron interacts with the atom and allows the bound electron to absorb, and not just scatter, the photon).
     
    Last edited: Jan 4, 2012
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